3.798 \(\int \frac{\sqrt [4]{a-b x^2}}{x^6} \, dx\)

Optimal. Leaf size=128 \[ -\frac{b^{5/2} \left (1-\frac{b x^2}{a}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{12 a^{3/2} \left (a-b x^2\right )^{3/4}}+\frac{b^2 \sqrt [4]{a-b x^2}}{12 a^2 x}+\frac{b \sqrt [4]{a-b x^2}}{30 a x^3}-\frac{\sqrt [4]{a-b x^2}}{5 x^5} \]

[Out]

-(a - b*x^2)^(1/4)/(5*x^5) + (b*(a - b*x^2)^(1/4))/(30*a*x^3) + (b^2*(a - b*x^2)^(1/4))/(12*a^2*x) - (b^(5/2)*
(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(12*a^(3/2)*(a - b*x^2)^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0455706, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {277, 325, 233, 232} \[ \frac{b^2 \sqrt [4]{a-b x^2}}{12 a^2 x}-\frac{b^{5/2} \left (1-\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{12 a^{3/2} \left (a-b x^2\right )^{3/4}}+\frac{b \sqrt [4]{a-b x^2}}{30 a x^3}-\frac{\sqrt [4]{a-b x^2}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(1/4)/x^6,x]

[Out]

-(a - b*x^2)^(1/4)/(5*x^5) + (b*(a - b*x^2)^(1/4))/(30*a*x^3) + (b^2*(a - b*x^2)^(1/4))/(12*a^2*x) - (b^(5/2)*
(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(12*a^(3/2)*(a - b*x^2)^(3/4))

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int \frac{\sqrt [4]{a-b x^2}}{x^6} \, dx &=-\frac{\sqrt [4]{a-b x^2}}{5 x^5}-\frac{1}{10} b \int \frac{1}{x^4 \left (a-b x^2\right )^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{a-b x^2}}{5 x^5}+\frac{b \sqrt [4]{a-b x^2}}{30 a x^3}-\frac{b^2 \int \frac{1}{x^2 \left (a-b x^2\right )^{3/4}} \, dx}{12 a}\\ &=-\frac{\sqrt [4]{a-b x^2}}{5 x^5}+\frac{b \sqrt [4]{a-b x^2}}{30 a x^3}+\frac{b^2 \sqrt [4]{a-b x^2}}{12 a^2 x}-\frac{b^3 \int \frac{1}{\left (a-b x^2\right )^{3/4}} \, dx}{24 a^2}\\ &=-\frac{\sqrt [4]{a-b x^2}}{5 x^5}+\frac{b \sqrt [4]{a-b x^2}}{30 a x^3}+\frac{b^2 \sqrt [4]{a-b x^2}}{12 a^2 x}-\frac{\left (b^3 \left (1-\frac{b x^2}{a}\right )^{3/4}\right ) \int \frac{1}{\left (1-\frac{b x^2}{a}\right )^{3/4}} \, dx}{24 a^2 \left (a-b x^2\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a-b x^2}}{5 x^5}+\frac{b \sqrt [4]{a-b x^2}}{30 a x^3}+\frac{b^2 \sqrt [4]{a-b x^2}}{12 a^2 x}-\frac{b^{5/2} \left (1-\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{12 a^{3/2} \left (a-b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0103868, size = 52, normalized size = 0.41 \[ -\frac{\sqrt [4]{a-b x^2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{4};-\frac{3}{2};\frac{b x^2}{a}\right )}{5 x^5 \sqrt [4]{1-\frac{b x^2}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(1/4)/x^6,x]

[Out]

-((a - b*x^2)^(1/4)*Hypergeometric2F1[-5/2, -1/4, -3/2, (b*x^2)/a])/(5*x^5*(1 - (b*x^2)/a)^(1/4))

________________________________________________________________________________________

Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{6}}\sqrt [4]{-b{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^2+a)^(1/4)/x^6,x)

[Out]

int((-b*x^2+a)^(1/4)/x^6,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b x^{2} + a\right )}^{\frac{1}{4}}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^2+a)^(1/4)/x^6,x, algorithm="maxima")

[Out]

integrate((-b*x^2 + a)^(1/4)/x^6, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b x^{2} + a\right )}^{\frac{1}{4}}}{x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^2+a)^(1/4)/x^6,x, algorithm="fricas")

[Out]

integral((-b*x^2 + a)^(1/4)/x^6, x)

________________________________________________________________________________________

Sympy [C]  time = 1.19517, size = 36, normalized size = 0.28 \begin{align*} - \frac{\sqrt [4]{a}{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, - \frac{1}{4} \\ - \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{5 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**2+a)**(1/4)/x**6,x)

[Out]

-a**(1/4)*hyper((-5/2, -1/4), (-3/2,), b*x**2*exp_polar(2*I*pi)/a)/(5*x**5)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b x^{2} + a\right )}^{\frac{1}{4}}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^2+a)^(1/4)/x^6,x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(1/4)/x^6, x)